Independent solution

How to solve this Exponential Distribution question

Answer in brief

This is a shifted-exponential percentile calculation. Setting the CDF to 0.10 gives a standardized excess of log(10/9), so the percentile is d+beta log(10/9), which is choice D.

Setup

Setup

Express the supported variable as its lower endpoint plus an exponential excess with scale beta.

X=d+Y,YExp(scale β)X=d+Y,\qquad Y\sim\operatorname{Exp}(\text{scale }\beta)
FX(x)=1e(xd)/β,xdF_X(x)=1-e^{-(x-d)/\beta},\qquad x\ge d

Model

Model

Let q denote the 10th percentile and set the shifted-exponential CDF at q equal to 0.10.

1e(qd)/β=0.101-e^{-(q-d)/\beta}=0.10
e(qd)/β=0.90e^{-(q-d)/\beta}=0.90

Compute

Compute

Take logarithms and solve for the percentile on the original scale.

qdβ=log(0.90)=log ⁣(109)\frac{q-d}{\beta}=-\log(0.90)=\log\!\left(\frac{10}{9}\right)
q=d+βlog ⁣(109)q=d+\beta\log\!\left(\frac{10}{9}\right)

Answer

Answer

The fixed shift must remain in the percentile, and beta multiplies the dimensionless logarithm.

q0.10=d+βlog(10/9)(D)\boxed{q_{0.10}=d+\beta\log(10/9)\quad\text{(D)}}